Find: 30 C mg L θd=1.047 Kd,20 C=0.11 [day-1]

Find: BOD5-day @ 30 C mg L θd=1.047 Kd,20 C=0.11 [day-1]
DO0 days = (saturated) mg DO5 days = 5.7 L time mg chloride = 0 140 170 210 240 Find the biochemical oxygen demand after 5 days, at 30 degrees Celsius, in milligrams per liter. [pause] In this problem, --- L dilution: T=20 C o sample = 10 [ml] water = 490 [ml]

DO0 days = (saturated) mg DO5 days = 5.7 L time mg chloride = 0 140 170 210 240 10 milliliters of a sample, is diluted, with 490 milliliters of water, at 20 degrees Celsius. L dilution: T=20 C o sample = 10 [ml] water = 490 [ml]

DO0 days = (saturated) mg DO5 days = 5.7 L time mg chloride = 0 140 170 210 240 Initially, the solution is saturated with dissolved oxygen, and after 5 days, the dissolved oxygen measures 5.7 milligrams per liter. L dilution: T=20 C o sample = 10 [ml] water = 490 [ml]

DO0 days = (saturated) mg DO5 days = 5.7 L time mg chloride = 0 140 170 210 240 There is no chloride in the sample, and the temperature variation constant and the deoxygenation rate constant for 20 degrees Celsius are provided. L dilution: T=20 C o sample = 10 [ml] water = 490 [ml]

Find: BOD5-day @ 30 C mg L θd=1.047 chloride = 0 Kd,20 C=0.11 [day-1]
dilution 20 C o DO0 days,20 C = (saturated) o sample = 10 [ml] mg L DO5 days,20 C = 5.7 o water = 490 [ml] BODt = BODu * (1-10-kd * t) The biochemical oxygen demand at a time, t, in milligrams per liter, equals, --- mg L biochemical oxygen demand

dilution 20 C o DO0 days,20 C = (saturated) o sample = 10 [ml] mg L DO5 days,20 C = 5.7 o water = 490 [ml] BODt = BODu * (1-10-kd * t) the ultimate biochemical oxygen demand, BOD sub u, in milligrams per liter, times the quantity, 1 minus ---- mg L ultimate biochemical oxygen demand mg biochemical oxygen demand L

dilution 20 C o DO0 days,20 C = (saturated) o sample = 10 [ml] mg L DO5 days,20 C = 5.7 o water = 490 [ml] BODt = BODu * (1-10-kd * t) 10 raised to –1 times the deoxygenation rate constant, k sub d, in unit of 1 over days, --- deoxygen rate constant [day-1] mg ultimate biochemical oxygen demand L mg biochemical oxygen demand L

dilution 20 C o DO0 days,20 C = (saturated) o sample = 10 [ml] mg L DO5 days,20 C = 5.7 o water = 490 [ml] BODt = BODu * (1-10-kd * t) time [day] times the time, t, in days. [pause] From the problem statement, we know to use a time of --- deoxygen rate constant [day-1] mg ultimate biochemical oxygen demand L mg biochemical oxygen demand L

dilution 20 C o DO0 days,20 C = (saturated) o sample = 10 [ml] mg L DO5 days,20 C = 5.7 o water = 490 [ml] BODt = BODu * (1-10-kd * t) time [day] 5 days, but we don’t yet know the ultimate BOD, --- deoxygen rate constant [day-1] mg ultimate biochemical oxygen demand L mg biochemical oxygen demand L

Find: BOD5-day @ 30 C ? ? mg L θd=1.047 chloride = 0
Kd,20 C=0.11 [day-1] o dilution 20 C o DO0 days,20 C = (saturated) o sample = 10 [ml] mg L DO5 days,20 C = 5.7 o water = 490 [ml] ? ? BODt = BODu * (1-10-kd * t) time [day] or the deoxygenation rate constant, for 30 degrees Celsius. We’ll begin by solving for the --- deoxygen rate constant [day-1] mg ultimate biochemical oxygen demand L mg biochemical oxygen demand L

Kd,20 C=0.11 [day-1] o dilution 20 C o DO0 days,20 C = (saturated) o sample = 10 [ml] mg L DO5 days,20 C = 5.7 o water = 490 [ml] ? ? BODt = BODu * (1-10-kd * t) time [day] ultimate BOD. The ultimate BOD demand equals --- deoxygen rate constant [day-1] mg ultimate biochemical oxygen demand L mg biochemical oxygen demand L

dilution 20 C o DO0 days,20 C = (saturated) o sample = 10 [ml] mg DO5 days,20 C = 5.7 o water = 490 [ml] L BODt = BODu * (1-10-kd * t) the BOD at a given time period, divided by the percentage change between initial and ultimate BOD, up to that time, as a decimal. From the plot of, --- BODt BODu = (1-10-kd * t)

dilution 20 C o DO0 days,20 C = (saturated) o sample = 10 [ml] mg DO5 days,20 C = 5.7 o water = 490 [ml] L BOD BODU BODt BODu = BOD vs. time, we notice the ultimate BOD is temperature independent, which means, the ultimate BOD, at 20 degrees Celsius, equals the ultimate BOD at 30 degrees Celsius, and --- T=30 C o (1-10-kd * t) T=20 C o time

dilution 20 C o DO0 days,20 C = (saturated) o sample = 10 [ml] mg DO5 days,20 C = 5.7 o water = 490 [ml] L BOD BODU BODt BODu = we can solve for the right hand side of the equation for any temperature. Using 20 degrees Celsius, we know the --- T=30 C o (1-10-kd * t) T=20 C o time

dilution 20 C o DO0 days,20 C = (saturated) o sample = 10 [ml] mg DO5 days,20 C = 5.7 o water = 490 [ml] L BOD BODU BODt BODu = deoxygenate rate constant, k sub d, but we don’t yet know the --- T=30 C o (1-10-kd * t) T=20 C o time

Kd,20 C=0.11 [day-1] o dilution 20 C o DO0 days,20 C = (saturated) o sample = 10 [ml] mg DO5 days,20 C = 5.7 o water = 490 [ml] L BOD ? BODU BODt BODu = time, t, or the BOD at time t. [pause] Since we’ve been given data for the concentration of --- T=30 C o (1-10-kd * t) T=20 C o ? time

Kd,20 C=0.11 [day-1] o dilution 20 C o DO0 days,20 C = (saturated) o sample = 10 [ml] mg DO5 days,20 C = 5.7 o water = 490 [ml] L BOD ? BODU BODt BODu = dissolved oxygen, at 5 days, we’ll use the time, t, --- T=30 C o (1-10-kd * t) T=20 C o ? time

Find: BOD5-day @ 30 C ? mg L θd=1.047 chloride = 0
Kd,20 C=0.11 [day-1] o dilution 20 C o DO0 days,20 C = (saturated) o sample = 10 [ml] mg DO5 days,20 C = 5.7 o water = 490 [ml] L BOD ? BODU BODt BODu = of 5 days, and then solve for the BOD at time at 5 days. T=30 C o (1-10-kd * t) T=20 C o 5 days time

Kd,20 C=0.11 [day-1] o dilution 20 C o DO0 days,20 C = (saturated) o sample = 10 [ml] mg DO5 days,20 C = 5.7 o water = 490 [ml] L BOD ? BODU BODt BODu = The biochemical oxygen demand of a sample equals, --- T=30 C o (1-10-kd * t) T=20 C o 5 days time

Kd,20 C=0.11 [day-1] o dilution 20 C o DO0 days,20 C = (saturated) o sample = 10 [ml] mg DO5 days,20 C = 5.7 o water = 490 [ml] L ? BOD5 BODu = the change in dissolved oxygen concentration, of that sample, times the dilution factor, DF (1-10-kd * t) 5 days BOD5 = (DO0 days-DO5 days) * DF

Kd,20 C=0.11 [day-1] o dilution 20 C o DO0 days,20 C = (saturated) o sample = 10 [ml] mg DO5 days,20 C = 5.7 o water = 490 [ml] L Vtotal ? DF= BOD5 Valiquot BODu = The dilution factor equals, the total volume, divided by the aliquot volume, of the tested sample. (1-10-kd * t) 5 days BOD5 = (DO0 days-DO5 days) * DF

Kd,20 C=0.11 [day-1] o dilution 20 C o DO0 days,20 C = (saturated) o sample = 10 [ml] mg DO5 days,20 C = 5.7 o water = 490 [ml] L Vtotal (10+490) [ml] ? DF= = BOD5 Valiquot 10 [ml] BODu = Plugging in the given volumes for the sample and water, the dillution factor equals, --- (1-10-kd * t) 5 days BOD5 = (DO0 days-DO5 days) * DF

Kd,20 C=0.11 [day-1] o dilution 20 C o DO0 days,20 C = (saturated) o sample = 10 [ml] mg DO5 days,20 C = 5.7 o water = 490 [ml] L Vtotal (10+490) [ml] ? DF= = BOD5 Valiquot 10 [ml] BODu = 50. [pause] The problem states the concentration of dissolved oxygen at 5 days equals --- (1-10-kd * t) 5 days DF=50 BOD5 = (DO0 days-DO5 days) * DF

Kd,20 C=0.11 [day-1] o dilution 20 C o DO0 days,20 C = (saturated) o sample = 10 [ml] mg DO5 days,20 C = 5.7 water = 490 [ml] o L Vtotal (10+490) [ml] ? DF= = BOD5 Valiquot 10 [ml] BODu = 5.7 milligrams per liter, and that the initial sample is --- (1-10-kd * t) 5 days DF=50 BOD5 = (DO0 days-DO5 days) * DF

Kd,20 C=0.11 [day-1] o dilution 20 C o DO0 days,20 C = (saturated) o sample = 10 [ml] mg DO5 days,20 C = 5.7 water = 490 [ml] o L Vtotal (10+490) [ml] ? DF= = BOD5 Valiquot 10 [ml] BODu = saturated. The concentration of dissolved oxygen in a saturated sample, --- (1-10-kd * t) 5 days DF=50 BOD5 = (DO0 days-DO5 days) * DF

Kd,20 C=0.11 [day-1] o L dilution 20 C o DO0 days,20 C = (saturated) o sample = 10 [ml] mg DO5 days,20 C = 5.7 water = 490 [ml] o L Vtotal (10+490) [ml] ? DF= = BOD5 Valiquot 10 [ml] BODu = free of chloride, at 20 degrees Celsius, can be looked up in a table, and equals --- (1-10-kd * t) 5 days DF=50 BOD5 = (DO0 days-DO5 days) * DF

Kd,20 C=0.11 [day-1] o L dilution 20 C mg o DO0 days,20 C = 9.17 o sample = 10 [ml] L mg DO5 days,20 C = 5.7 water = 490 [ml] o L Vtotal (10+490) [ml] ? DF= = BOD5 Valiquot 10 [ml] BODu = 9.17, milligrams per liter. [pause] Next, the dissolved oxygen concentrations and dilution factor are --- (1-10-kd * t) 5 days DF=50 BOD5 = (DO0 days-DO5 days) * DF

Kd,20 C=0.11 [day-1] o L dilution 20 C mg o DO0 days,20 C = 9.17 o sample = 10 [ml] L mg DO5 days,20 C = 5.7 water = 490 [ml] o L Vtotal (10+490) [ml] ? DF= = BOD5 Valiquot 10 [ml] BODu = plugged into the equation, and the 5-day BOD at 20 degrees equals, --- (1-10-kd * t) 5 days DF=50 BOD5 = (DO0 days-DO5 days) * DF

Kd,20 C=0.11 [day-1] o L dilution 20 C mg o DO0 days,20 C = 9.17 o sample = 10 [ml] L mg DO5 days,20 C = 5.7 water = 490 [ml] o L Vtotal (10+490) [ml] ? DF= = BOD5 Valiquot 10 [ml] BODu = 173.5 milligrams per liter. [pause] (1-10-kd * t) 5 days DF=50 BOD5 = (DO0 days-DO5 days) * DF mg BOD5 = 173.5 L

dilution 20 C mg o DO0 days,20 C = 9.17 o sample = 10 [ml] L mg DO5 days,20 C = 5.7 water = 490 [ml] o L Vtotal mg (10+490) [ml] BOD5 = 173.5 DF= = L Valiquot 10 [ml] At this point we can return to our equation for the ultimate BOD, --- BOD5 DF=50 BODu = (1-10-kd * t)

dilution 20 C mg o DO0 days,20 C = 9.17 o sample = 10 [ml] L mg DO5 days,20 C = 5.7 water = 490 [ml] o L Vtotal mg (10+490) [ml] BOD5 = 173.5 DF= = L Valiquot 10 [ml] plug in our values for the 5-day BOD, the deoxygenate rate constant, and the time, and the ultimate BOD equals, --- BOD5 t=5 [days] DF=50 BODu = (1-10-kd * t)

dilution 20 C mg o DO0 days,20 C = 9.17 o sample = 10 [ml] L mg DO5 days,20 C = 5.7 water = 490 [ml] o L Vtotal mg (10+490) [ml] BOD5 = 173.5 DF= = L Valiquot 10 [ml] 241.6, milligrams per liter. [pause] Returning to our initial equation for the --- BOD5 t=5 [days] DF=50 BODu = (1-10-kd * t) mg BODu =241.6 L

dilution 20 C o DO0 days,20 C = (saturated) o sample = 10 [ml] mg L DO5 days,20 C = 5.7 o water = 490 [ml] mg BODu =241.6 L BODt = BODu * (1-10-kd * t) biochemical oxygen demand, we have the ultimate BOD ---

dilution 20 C o DO0 days,20 C = (saturated) o sample = 10 [ml] mg L DO5 days,20 C = 5.7 o water = 490 [ml] mg BODu =241.6 L BODt = BODu * (1-10-kd * t) and the time, t, as 5 days, but we need to determine the deoxygenation rate, ---

Kd,20 C=0.11 [day-1] o dilution 20 C o DO0 days,20 C = (saturated) o sample = 10 [ml] mg L DO5 days,20 C = 5.7 o water = 490 [ml] mg BODu =241.6 L BODt = BODu * (1-10-kd * t) k sub d, for a the temperature of 30 degrees. [pause] From the plot of BOD vs time, we notice--- ?

dilution 20 C o DO0 days,20 C = (saturated) o sample = 10 [ml] mg L DO5 days,20 C = 5.7 o water = 490 [ml] BOD BODt = BODu * (1-10-kd * t) the reaction at 30 degrees Celsius, is faster than the reaction at 20 degrees Celsius, so after a certain time period, --- T=30 C o T=20 C o time

dilution 20 C o DO0 days,20 C = (saturated) o sample = 10 [ml] mg L DO5 days,20 C = 5.7 o water = 490 [ml] BOD BODt, 30 C o there will be a higher biochemical oxygen demand in the sample with the higher temperature. BODt, 20 C o T=30 C o T=20 C o time

dilution 20 C o DO0 days,20 C = (saturated) o sample = 10 [ml] mg L DO5 days,20 C = 5.7 o water = 490 [ml] BOD BODt, 30 C o The deoxygenation rate constant, can be converted between two different temperatures, by knowing the ---- BODt, 20 C o T=30 C o Kd,T = Kd,T * θdT -T 1 2 T=20 C o time

dilution 20 C o DO0 days,20 C = (saturated) o sample = 10 [ml] mg L DO5 days,20 C = 5.7 o water = 490 [ml] BOD BODt, 30 C o he deoxygenation rate constant at one temperature, and the temperature variation constant, theta sub d. BODt, 20 C o T=30 C o Kd,T = Kd,T * θdT -T 1 2 T=20 C o time

dilution 20 C o DO0 days,20 C = (saturated) o sample = 10 [ml] mg L DO5 days,20 C = 5.7 o water = 490 [ml] BOD BODt, 30 C o The deoxygenation rate at 30 degrees Celsius equals, --- BODt, 20 C o T=30 C o Kd,T = Kd,T * θdT -T 1 2 T=20 C o Kd,30 C=0.11[day-1] *1.047(30-20) o time

dilution 20 C o DO0 days,20 C = (saturated) o sample = 10 [ml] mg L DO5 days,20 C = 5.7 o water = 490 [ml] BOD BODt, 30 C o 0.174, day to the minus 1. [pause] Returning to the original equation, --- BODt, 20 C o T=30 C o Kd,T = Kd,T * θdT -T 1 2 T=20 C o Kd,30 C=0.11[day-1] *1.047(30-20) o time Kd,30 C=0.174 [day-1] o

dilution 20 C o DO0 days,20 C = (saturated) o sample = 10 [ml] mg L DO5 days,20 C = 5.7 o water = 490 [ml] mg BODu =241.6 Kd,30 C=0.174 [day-1] o L The solved variables are plugged in, and the --- time = 5 [days] BOD5 30 C = BODu * (1-10-kd * t) o

dilution 20 C o DO0 days,20 C = (saturated) o sample = 10 [ml] mg L DO5 days,20 C = 5.7 o water = 490 [ml] mg BODu =241.6 Kd,30 C=0.174 [day-1] o L 5-day BOD at 30 degrees Celsius equals, --- time = 5 [days] BOD5 30 C = BODu * (1-10-kd * t) o

dilution 20 C o DO0 days,20 C = (saturated) o sample = 10 [ml] mg L DO5 days,20 C = 5.7 o water = 490 [ml] mg BODu =241.6 Kd,30 C=0.174 [day-1] o L 209, milligrams per liter. time = 5 [days] BOD5 30 C = BODu * (1-10-kd * t) o mg L BOD5, 30 C = 209 o

dilution 20 C o DO0 days,20 C = (saturated) o sample = 10 [ml] mg L DO5 days,20 C = 5.7 o water = 490 [ml] mg BODu =241.6 Kd,30 C=0.174 [day-1] o L 140 170 210 240 Looking over the possible solutions, --- time = 5 [days] BOD5 = BODu * (1-10-kd * t) mg L BOD5, 30 C = 209 o

dilution 20 C o DO0 days,20 C = (saturated) o sample = 10 [ml] mg L DO5 days,20 C = 5.7 o water = 490 [ml] mg BODu =241.6 Kd,30 C=0.174 [day-1] o L 140 170 210 240 the answer is C. time = 5 [days] BOD5 = BODu * (1-10-kd * t) mg L AnswerC BOD5, 30 C = 209 o

? Index σ’v = Σ γ d γT=100 [lb/ft3] +γclay dclay 1
Find: σ’v at d = 30 feet (1+wc)*γw wc+(1/SG) σ’v = Σ γ d d Sand 10 ft γT=100 [lb/ft3] 100 [lb/ft3] 10 [ft] 20 ft Clay = γsand dsand +γclay dclay A W S V [ft3] W [lb] 40 ft text wc = 37% ? Δh 20 [ft] (5 [cm])2 * π/4

Find: 30 C mg L θd=1.047 Kd,20 C=0.11 [day-1]

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Find: 30 C mg L θd=1.047 Kd,20 C=0.11 [day-1]

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Presentation on theme: "Find: 30 C mg L θd=1.047 Kd,20 C=0.11 [day-1]"— Presentation transcript:

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